once the integrals involved in our expression for V, in terms of the equation to the surface of the attracting mass, without integration.

From the expression for the attraction can be immediately found by the formula of Art. 20. Thus


dr 4πρα8 4πρα 2a

(i + 1) a + {Y. +

Yi+...}. 372


(2i + 1) gi PROP. To find the attraction of a homogeneous body, differing but little from a sphere, on a particle within its mass.

Y, +...

43. We must in this case expand V in an ascending series of powers of r; and shall therefore take the second of the series of Art. 40. By proceeding as in the last Proposition, we find that the part of V which appertains to the excess over the sphere

=ef S*{a?Pn+arP, +...+ Pi+...} y'du' dw,

} ,

= 4rpa" {Y.+ 3a Y, +...+ Yi+...}.

(2i+1) a Adding to this the part of V which appertains to the sphere of radius a, viz, 2mpa’ – Tepro, for the whole mass,

:poi V=2mpa? pro +47 pa{Y.+

Y, +...+


(2i + 1) a

dV And the attraction =





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pr – Ampa fj Y, +




Y, + ... + 5a

(2i+1) at1 We can show that by properly choosing the value of (a) and the origin of the radius of the surface we can make Y, and Y, disappear from the above formulæ.

PROP. To show that by choosing a equal to the radius of the sphere of which the mass equals that of the attracting body we cause Y to vanish, and by taking the centre of gravity of the body as the origin of the radius vector, we causé Ý, to vanish.

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where r is the radius vector of the surface of the body, and = a (1 + y) suppose. Putting this for r mass of body

= mass of sphere (rad. = a) +pa" | Soydudes

LT" Y., by Art. 26,

= mass of sphere + pas

= mass of sphere + 47 pa' Y.. If then a be taken equal to the radius of the sphere of which the mass equals the mass of the body, Yo=0, as was stated.

Again, let wyz be the co-ordinates to the centre of gravity of the body, M its mass: the co-ordinates to the element of which the mass is - pri drdu dw are PV1 - * cos w, rv 1 - på sin w, and


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OS"proudrdydw = 1S S*r*udydo; putting r=a (1 + y) = a (1 + Y.+Y, + ... + Y,+...), and observing that V1 - cos w, N1 - sin

u* sin w, and u satisfy Laplace's Equation, and are of the first order, we have by Art. 26,

M.q=pa" | S*Y, VI— j* cos ao dudo,
M.q=pa" | SY, V1– * sin wdyda,
M.a = pa


S* Y, ududa. But Y,, being a function of p, V1-rcos w, and N1-uʼsin w of the first order, is of the form

AN1 - cos w + BNI - på sin w+ Cpe; .. Μ.- πρα'A, M., πρωB, M.3 = πραα.

Hence if we take the origin of co-ordinates at the centre of gravity and therefore w=0, y=0, 2=0, we have A=0, B=0, C=0, and therefore Y, = 0, as stated in the enunciation.

PROP. To find the attraction of a heterogeneous body upon a particle without it; the body consisting of thin strata nearly spherical, homogeneous in themselves, but differing one from another in density.

45. Let a' (1+y') be the radius of the external surface of any stratum, a' being chosen so that

y' = Y/+Y+ ... + Yi + ... (Art. 44). Since the strata are supposed not to be similar to one another, y' is a function of a as well as of u and w'. Let é be the density of the stratum of which the mean radius is a'. Now the value of V for this stratum equals the difference between the values of V for two homogeneous bodies of Y +

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the density é and mean radii a' and a' – da'. But for the body of which the mean radius is a' (Art. 42)

47 p'a's , 4p'a' sa' V +


Y' +


(2i + 1) po Hence for the stratum of which the external mean radius is a', 4πρ? 4πρ' d (α4

a' V=


+ g da (3r 2i+mai and therefore for the whole body, 47

d V=

Y; +

Y +
da' 3r


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li+3 a

so fan


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(2i+1) put

lits a

da 3r

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From which the attraction is easily deduced.

PROP. To find the attraction of the same body on an internal particle.

46. Let r=a (1+y) be the radius of the stratum in which the attracted particle lies. Then for the strata within the surface of which the radius is a (1+y), we have 47

d la'
P +

Y' +



(21 + 1) 7 But for a stratum external to the particle we have by Art. 43,

d V=47p'a'da' +47p'


Y/' +...+ da' 3

1) a"Consequently for the whole body, 47

d V: +


da' da (3r

) d fra

get + 47 p



(2ias From this the attraction is readily obtained by differentiating with respect to r,







47. THE methods which have hitherto been given enable us to find the attraction of the Earth and other bodies of our system considered as a whole. But, taking the Earth as our example, the surface is irregular, and neither exactly spherical nor spheroidal. We ought, therefore, to be able to calculate the effect of these irregularities, and with this view the present Chapter is added to what has gone before. High Table-lands may very materially affect the position of the plumb-line in some places. Enormous irregular mountain masses, like the Himmalayas, may do the same. Their effect ought, therefore, to be carefully estimated, as all instruments which are fixed by the plumb-line or spirit-level must be affected by such irregularities.

PROP. To find the attraction of a slender prism of matter on a point in the line drawn to one of its extremities.

48. Let AB be the prism, C the attracted point, P any element of the prism, AP=r,

M the mass and l the length
of the prism, AC=a, BC=b,
PC=y, angle PAC=0.
Then the mass of the ele-

dr ment at P=M

ī: Attraction of element at P .dr 1

P Р on C= M

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